SillyOldDuffer | 17/08/2017 17:29:14 |
10668 forum posts 2415 photos | Andrew Tinsley started a thread on Dore Small Boring Bars that produced comments on the type of boring bar where a push-rod or fluid is used to tension the bar. This is said to improve rigidity. True or Bunkum? Using a 750mm length of 1/2" gas pipe and some 8mm studding I set up an experiment for a real world test. The results showed 'Bunkum' to be the winner. Unfortunately Michael Gilligan (on top critical form after his holiday) pointed out that I'd compressed the rod rather than stretching it. I was busted! Here's the second attempt, this time - I hope - doing it right. I put an M13.7 x 1.25 thread on the end of the pipe. Then I turned and threaded a blind collar to screw on to the pipe. (Metric thread on an imperial diameter, yay!) Then I drilled and tapped a hole through the collar to take the M8 studding. I blocked the other end of the pipe with a top-hat pinned in place, plus a greased steel bearing pad. The parts can be seen on the right in the photo above. The pin is 4.5mm diameter mild steel. The pin is bent because the photo was taken after the experiment. The hole in the top hat is off centre due to bad workmanship. But it doesn't look too awful and nothing broke. I cut flats on the end of the studding and used a mole wrench to tighten it thus tensioning the pipe. Just like last time I stuck a pointer to the end of the pipe to measure movement. Then with one end of the pipe clamped in my vice I was able to tension the pipe by tightening the studding. During the test a kilogram of old springs were hung on the end of the pipe. The results were 'quite interesting'. Firstly, tightening the studding doesn't move the pointer much; it dropped a mm or two and after that more tightening didn't make much difference. (Apart from a few interesting 'metal complaining' noises) Secondly, it does seem that tensioning the pipe reduces vibration by nearly half. To test this I twanged the pipe sideways and measured the time it took to come to a visible rest with a stopwatch. Push rod hand tight - 2.25 seconds Push rod plus 1/2 turn - 1.41 seconds Push rod plus 1 turn - 1.35 seconds Push rod plus 1½ turns - 1.31 seconds Note the improvement isn't linear with increasing tension. So not bunkum. I'm not convinced the improvement is worth the effort though. Possibly a tool-steel would perform better than the nasty gas-pipe mild-steel I have in my junk box. All comments welcome. Won't be surprised to find I've cocked up again... Got a few unanswered questions out of doing this. For example, what's the optimum diameter and position of the retaining pin at the top-hat end? Too small and it will shear; too large and the hole in the pipe will give way. Dave Edited By SillyOldDuffer on 17/08/2017 17:29:55 |
Michael Gilligan | 17/08/2017 18:06:39 |
![]() 23121 forum posts 1360 photos | Nice work, Dave ... This time I think you're onto something. I knew we could rely on you. MichaelG. . P.S. ... Your results indicate increased damping, which is (a) useful, and (b) reasonably predictable I'm not yet convinced to accept Hemingway's "surprisingly stiff" claim though. Increased stiffness should increase the 'twang' frequency. Edited By Michael Gilligan on 17/08/2017 18:11:25 |
jason udall | 17/08/2017 18:45:14 |
2032 forum posts 41 photos | Just a thought. Musicians adjust the tension in wires. ..to change the fundamental frequency. If one were to get the note wrong. .you might get harmonically related vibration from tool and part. .being able to tune the tool to avoid this combination might prove useful. . Btw have used solid tungsten carbide ground tools..these do the job when all else fails. .seems "solid" tc has a lower Q than steel |
Muzzer | 17/08/2017 19:42:08 |
![]() 2904 forum posts 448 photos | Surely your "control" sample should be a solid bar? You would then compare that to a solid bar that has been drilled out and had a rod stuck firmly up its jaxi - assuming you have managed to couple them together either in compression or tension, with a force greater than you will subject it to as a result of your "test" force. Perhaps I'm missing something. Otherwise, any rod inside a pipe will obviously stiffen it. Murray |
JasonB | 17/08/2017 20:29:14 |
![]() 25215 forum posts 3105 photos 1 articles | Might also be interesting to measure deflection when a weight is hung on the end of the plain pipe and then when under various tensions as although the bar may chatter less it won't be much good if it deflects and gives a tapered bore. Also interesting to see what it would be like with a couple of spacer rings along the studding to keep it central as being free within the bore it may bow when compressed much like the tube did. Thanks for taking the effort to try things out |
SillyOldDuffer | 17/08/2017 20:46:27 |
10668 forum posts 2415 photos | Hi Murray, Good point about using a solid bar as a control. Wouldn't my pipe with it's pushrod hand tight be similar to a solid bar? Probably not, there's no cohesion between the side of the pipe and the side of the rod. Sorry, I don't follow your jaxi suggestion. Do you mean in the way that might be achieved by allowing a hot tube to cool on a cold mandrel whilst the two are pinned together at the ends? Anyway, isn't it enough to show that a pipe and compressed push-rod combi are stiffer than a solid rod of the same size? That question's, not been answered yet! Could this be simulated with Fusion360? I've had a play with FEM but that kind of analysis is well outside my comfort zone. Aside from drawing it, I wouldn't know where to start with this problem. Dave
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KWIL | 17/08/2017 20:53:51 |
3681 forum posts 70 photos | When the expected conclusion is finally drawn and yes the tensioned tube is stiffer (and therefore a different resonant frequency), just wind the VFD down or up and also reduce the chatter. |
SillyOldDuffer | 17/08/2017 21:13:24 |
10668 forum posts 2415 photos | Posted by JasonB on 17/08/2017 20:29:14:
Might also be interesting to measure deflection when a weight is hung on the end of the plain pipe and then when under various tensions as although the bar may chatter less it won't be much good if it deflects and gives a tapered bore. Also interesting to see what it would be like with a couple of spacer rings along the studding to keep it central as being free within the bore it may bow when compressed much like the tube did. Thanks for taking the effort to try things out Hi Jason, I did measure pipe deflection under load and with different tensions. Adding tension doesn't make much difference to the deflection, at least not in my arrangement. Spacer rings are a good idea. The bore of the pipe is a bit over 8.5mm, so 8mm studding has wiggle room. Muzzer got me thinking about how I might bore a straight 8mm diameter hole through 750mm of 1/2" steel-rod. Bit of a challenge as I've never drilled a hole in metal more than about 100mm deep. As for taking the effort I'm starting to think I've bitten off more than I can chew. Designing the test rig was mostly guesswork, for example the collar and pipe are held together with 12 turns of thread 1.2mm deep. Is that massive overkill or am I lucky nothing broke? Would the collar have been strong enough if I'd made it in Aluminium? Making the test piece was good fun basic training, thinking about how to design it properly is making my head hurt. Cheers, Dave |
DrDave | 17/08/2017 21:45:43 |
264 forum posts 52 photos | That is a very useful set of tests that you have performed, SillyOldDuffer, and you have made a few grey cells twitch. There has been lots of effort put in to try to understand what is going on with a preloaded boring bar. But, the stiffness of a boring bar is purely a function of its second moment of area. So why do tests appear to show otherwise? Duncan Webster is right with his post in the other thread (Dore Small Boring Bars): a solid bar will have the same stiffness as a hollow bar with a close-fitting “pusher”, independent of any end load. In real life, however, the pusher will be a rattle fit in the bar, rather than a tight fit. As a result, the pusher cannot contribute to bending stiffness until the outer bar has deflected enough to touch it. It can take significant load before the two start to act together so the hollow boring bar is indeed not as stiff as the solid one. Preloading the pusher and bar forces them to start to work together, even though there is a gap between them, effectively stiffening them up. But a hollow bar with push rod can NEVER be stiffer than a solid bar of the same outside diameter, no matter how tight they are tightened. Simply, the laws of nature dictate that that is the case. DrDave |
David Standing 1 | 17/08/2017 21:47:56 |
1297 forum posts 50 photos | Posted by SillyOldDuffer on 17/08/2017 21:13:24:
Muzzer got me thinking about how I might bore a straight 8mm diameter hole through 750mm of 1/2" steel-rod. Cheers, Dave
What you need is a tensioned boring bar |
Michael Gilligan | 17/08/2017 23:04:14 |
![]() 23121 forum posts 1360 photos | Posted by DrDave on 17/08/2017 21:45:43:
That is a very useful set of tests that you have performed, SillyOldDuffer, and you have made a few grey cells twitch. There has been lots of effort put in to try to understand what is going on with a preloaded boring bar. But, the stiffness of a boring bar is purely a function of its second moment of area. So why do tests appear to show otherwise? [ ... ] But a hollow bar with push rod can NEVER be stiffer than a solid bar of the same outside diameter, no matter how tight they are tightened. Simply, the laws of nature dictate that that is the case. DrDave . DrDave, On the face of it, you are right of course ... But I have a simple hypothesis which might merit consideration: The boring bar is [to a first approximation] a simple cantilever beam which is subjected to downward force applied at the unsupported end. Bending of the bar induces compression at the bottom of the beam, and tension at its top [easier to visualise with a rectagular beam]. If the bar is tensioned by the pushrod then, I suggest, we have a situation comparable [albeit opposite] to a pre-loaded spring, such as we see on motor vehicle suspension. The effect of this pre-loading would, I think, be to effectively stiffen the bar; because the tip-load has first to overcome the pre-load before it can induce any movement. I await (a) ridicule, (b) mathematical analysis, or (c) experiment from those suitably qualified. MichaelG. |
duncan webster | 17/08/2017 23:34:03 |
5307 forum posts 83 photos | Posted by DrDave on 17/08/2017 21:45:43:
That is a very useful set of tests that you have performed, SillyOldDuffer, and you have made a few grey cells twitch. There has been lots of effort put in to try to understand what is going on with a preloaded boring bar. But, the stiffness of a boring bar is purely a function of its second moment of area. So why do tests appear to show otherwise? Duncan Webster is right with his post in the other thread (Dore Small Boring Bars): a solid bar will have the same stiffness as a hollow bar with a close-fitting “pusher”, independent of any end load. In real life, however, the pusher will be a rattle fit in the bar, rather than a tight fit. As a result, the pusher cannot contribute to bending stiffness until the outer bar has deflected enough to touch it. It can take significant load before the two start to act together so the hollow boring bar is indeed not as stiff as the solid one. Preloading the pusher and bar forces them to start to work together, even though there is a gap between them, effectively stiffening them up. But a hollow bar with push rod can NEVER be stiffer than a solid bar of the same outside diameter, no matter how tight they are tightened. Simply, the laws of nature dictate that that is the case. DrDave On further reflection I think there is a mechanism by which a loose push rod located radially but not angularly (ie pin joint) at each end can stiffen a tube. Imagine a tube built in at one end, initially horizontal with a side load. It will have an end slope, call it A. The loose fitting push rod is still straight, and so its slope is less, call it B. Call the tensile force in the tube T and the compressive force in the push rod P. As there is no external force in the horizontal direction, resolving horizontal. T*cosA = P*cosB, so T = P*cosB/cosA as A > B then cosB/cosA > 1 and T > P Now resolve T and P vertical and we get an upward force of T*sinA - P* sinB we've already seen that T > P and sinA > sinB, so this force is non zero, and acts to resist the side load. Whether this amounts very much I don't know, and it would be quite a sum to work it out properly. I have a feeling it is very small as the angles A & B are very small. Another way of looking at it is to think of the loose push rod moving off centre in the tube as the tube bends. This will increase the second moment of area of the assemby compared with the push rod being central, but it can't increase it to more than it would have been with a close fitting rod. I think that a close fitting push rod combination has the same second moment of area as a solid bar, but there could well be some damping advantage If a loose fitting push rod were so slender it buckled then all bets would be off! |
Robert Dodds | 17/08/2017 23:58:10 |
324 forum posts 63 photos | Michael G, Good to see you back. I can't agree completely with your analogy to a pre tensioned car spring. The load direction of our tool bar is at right angles to the preload whereas the car spring restraint is coincidental with the load direction . Bob D |
jimmy b | 18/08/2017 07:22:53 |
![]() 857 forum posts 45 photos | Impressive work Dave, well done.
Its a subject that can be gotten into too deep! If chatter/vibration is happening, there are so many things that can be done to eliminate it. The machines i run at work can vary the spindle speed, over a given range and time, i.e. +/-200rpm variation, (say 1600 to 2000 rpm) over two seconds, for example.
Another trick i have used on a lathe and on milling boring head is to put a lump of placticine on the tool. |
Neil Wyatt | 18/08/2017 07:44:47 |
![]() 19226 forum posts 749 photos 86 articles | Lets take the extreme case as a thought experiment. A thin flexible plastic tube, it just flops around. A whippy bit of piano wire. Plug the tube, thread the wire up it and push a bit. Is the end result stiffer than either the wire or tube alone? - Yes, it can't be less stiff than the sum of the two individually. Is the end results stiffer than the sum of the two alone? |
Neil Wyatt | 18/08/2017 08:08:53 |
![]() 19226 forum posts 749 photos 86 articles | Oh I'm an oaf. Over the last few month's I've been discussing 'tensegrity structures' (Word coined by Buckminster Fuller) with an architect who has designed them and I've just realised this is an example. Google 'an introduction to the mathematics of tensegrity'. Imagine a slender flagpole held erect by four guy ropes. It is far stiffer than pole or ropes alone. Our boring bar is a less ideal structure (a conical bar would work best!) but is still a tensegrity structure and yes it is increasing stiffness through preload, whether or not the amount is useful is debatable. Neil |
SillyOldDuffer | 18/08/2017 08:43:40 |
10668 forum posts 2415 photos | Posted by Neil Wyatt on 18/08/2017 07:44:47:
Lets take the extreme case as a thought experiment. A thin flexible plastic tube, it just flops around. A whippy bit of piano wire. Plug the tube, thread the wire up it and push a bit. Is the end result stiffer than either the wire or tube alone? - Yes, it can't be less stiff than the sum of the two individually. Is the end results stiffer than the sum of the two alone? Warning adult content follows! I was pondering last night about the 'Second Moment of Area', and whether or not it is the sole arbiter of deflection. (In lay terms I understand the Second Moment of Area to be a mathematical construct that makes allowance for the cross-sectional shape of the bar. That is square, rectangular, round, 'T' section and 'I' girders all behave differently, and the effect may be directional. For example a 12" steel rule bends much more up and down than it does right to left.) I'm not at all clear what 'Second Moment of Area' means in the context of a tensioned or pressurized pipe because another significant force has been introduced. Last night I was thinking nostalgically (as I often do) about the male intromittent organ. Naturally I mustn't refer to personal experience, but I understand this normally floppy tool is stiffened for use by the application of hydraulic pressure. Neil asked of piano wire pushed into a flexible tube: 'Is the end result stiffer than the sum of the two alone?' I'd say the answer in the case of a gentleman's willy is yes. The combination of an elastic tube and fluid under pressure behaves very differently from the tube and fluid apart! Bottom line at the moment is that I'm thoroughly confused. Apologies that I don't have time at the moment to give the other posts the consideration they deserve. I have to get outside and unblock a downpipe before it rains again. It's clouding over rapidly here! Dave PS Started raining now. Hard. Arghh!
Edited By SillyOldDuffer on 18/08/2017 08:44:07 Edited By SillyOldDuffer on 18/08/2017 08:46:29 |
Martin Kyte | 18/08/2017 09:52:12 |
![]() 3445 forum posts 62 photos | Not convinced Niel. Ideally the two comparisons for the boring bar are 1. the unstressed bar and 2. a composite bar where the 'core' is in tension and the tubular 'shell' is in compression. As it's a thought experiment the second bar can be a single peice of steel as we don't have to actually make it. If the elastic properties of the steel are identical in compression and tension I cannot see a mechanism by which bar 2 can be stiffer to a static bending moment than bar 1. I can however understand that the prestressed bar could have different properties with respect to the dynamic transmission of shock waves in the same way that a musical string changes pitch under different stresses. (as has been mensioned earlier). I have done some searching around on the web and have found patents for boring bars utilising carbide rings to create a more rigid outer tube and an inner core arranged to be in tension so that the outer rings are always in compression. A converse scheme where the outer tube is in tension could be practical if a harder material could be suitably employed. I suspect that the "stiffeness" of the bar we were originally discussing when this thread opened was much more related to the change in resonant frequency and does not affect the difflection under static load. If you really want an analog you could do worse than to think of a central coil spring surrounded by a number of similar springs all of which are attached to a disc at the top end and similarly at the bottom with the exception of the central spring which is attached to a screw passing through a tapped hole in the bottom disc . This allows the system to be either unstressed or stressed by a varying ammount. I seems intuitive that the resonant frequency (off axis oscillatio) would increase when the screw is tightened but again I fail to see a mechanism which makes the system any stiffer to a side force at the top. It's an intriguing puzzle and I am happy to be shown I am wrong, but as I said so far I am not convinced. regards Martin
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Neil Wyatt | 18/08/2017 09:52:16 |
![]() 19226 forum posts 749 photos 86 articles | I can't believe he posted that while (just) managing to avoud breaking forum rules... |
David Standing 1 | 18/08/2017 10:21:34 |
1297 forum posts 50 photos | Posted by Neil Wyatt on 18/08/2017 09:52:16:
I can't believe he posted that while (just) managing to avoud breaking forum rules...
It definitely has to be 'Post of the week' |
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